Let the maximum value of the expression $y=\frac{x^4-x^2}{x^6+2x^3-1}$ for $x>1$ is $\frac p q$,where $p$ and $q$ are relatively prime positive integers.Find $p+q$.

My attempt: $y=\frac{x^2(x^2-1)}{(x^3+1)^2-2}$, then I could not think of any way to find maximum value of $f(x)$. Can someone help me in solving for $p$ and $q$?


Let $t:=x-\frac{1}{x}$, which can take any value in $\mathbb{R}_{>0}$ Then, $y=\frac{t}{t^3+3t+2}$. To maximize $y$, we have to minimize $\frac{1}{y}=t^2+3+\frac{2}{t}$. By the AM-GM Inequality, $$\frac{1}{y}=\left(t^2+\frac{1}{t}+\frac{1}{t}\right)+3\geq 3\sqrt[3]{t^2\cdot\frac{1}{t}+\frac{1}{t}}+3=6\,.$$ That is, $y \leq \frac{1}{6}$. The equality holds if and only if $t^2=\frac{1}{t}$, or $t=1$. Hence, the maximum of $y$ is $\frac{1}{6}$, which is attained at $x=\frac{1+\sqrt{5}}{2}$.

  • $\begingroup$ Could you explain $$ y=\frac{t}{t^3+3t+2} $$ in detail? $\endgroup$ – user64494 Jul 31 '15 at 17:10
  • $\begingroup$ Divide the numerator and the denominator of the original expression of $y$ by $x^3$. $\endgroup$ – Batominovski Jul 31 '15 at 17:11
  • $\begingroup$ $$ { \left( x-{x}^{-1} \right) \left( {x}^{3}+2-{x}^{-3} \right) ^{-1}} $$ is obtained. What is next? $\endgroup$ – user64494 Jul 31 '15 at 17:15
  • $\begingroup$ What is $\left(x-\frac1x\right)^3$? $\endgroup$ – Batominovski Jul 31 '15 at 17:18

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